Potential difference Calculation

[Tan_Can]

I am stuck on this question about potential difference, the question is: each side of an equilateral triangle measures 3.61cm. A charge of -0.34mC is placed at each vertex. what is the potential difference at the midpoint of the base of the triangle, i need quick help. Thank you.

 

[basePARTICLE]

Sum the potentials arising from the charges, after drawing the vector diagram.

 

[Moetasim]

The potential difference at the midpoint of the base of the triangle can be calculated by using the formula V = kq/r, where V is the potential difference, k is the Coulomb's constant (9x10^9 Nm^2/C^2), q is the charge (-0.34mC), and r is the distance from the charge to the midpoint of the base.

First, we need to find the distance from the charge to the midpoint of the base. Since the triangle is equilateral, the distance from each vertex to the midpoint of the base is half of the length of the side, which is 3.61/2 = 1.805 cm.

Now, we can plug in the values into the formula:

V = (9x10^9 Nm^2/C^2)(-0.34x10^-3 C)/(1.805x10^-2 m) = -5.02 V

Therefore, the potential difference at the midpoint of the base is -5.02 volts. This means that a positive charge placed at this point would experience a force of 5.02 volts pushing it towards the negative charges at the vertices of the triangle. I hope this helps!